I am using this version of L'Hopital's rule

Assume that $\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a}g(x)=0$, and that the limit-value $\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$ exists (could be $\infty$ or $-\infty$). Then the limit value $\lim_{x \rightarrow a} \frac{f(x)}{g(x)}$ also exsits, and $\lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$.

I have two questions:

  1. When we have said that $\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$, exists, then we have also said that $g'(x)$ is not zero around $a$ , so this does not create a problem?

  2. This is my main question: Could there be a special case, where the hypothesis of the rule is satisfied, but for every neighbourhood around $a$ , there is an $x$ such that $g(x)=0$, and so that $\lim_{x \rightarrow a} \frac{f(x)}{g(x)}$ is not defined, and hence the rule stated as it is, is wrong? Because the rule says that the limit value of $\lim_{x \rightarrow a} \frac{f(x)}{g(x)}$ exists if the hypothesis is satisfied? (The reason I am suspecting this is what I write below about Cauchy's Mean Value Theorem).

On Wikipedia: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule they assume that $g'(x)$ is not zero. But they do not say anything about $g(x)$, how do we know that there are not infinitely many points around $a$ , where $g$ is zero?

The reason I am asking is that they use Cauchy's mean value theorem in my book for the proof. http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem But when using the form with fractions, they have to assume that the denominator is not zero? It doesn't say that if $g'(c)$ is non-zero then $g(b)-g(a)$ is nonzero, we are only allowed to go to the fraction part of the theorem if we know that both is non-zero?

  • $\begingroup$ why should 1. be a problem? $\endgroup$ – Loreno Heer Jan 29 '15 at 15:54
  • $\begingroup$ @sanjab I am not saying if it is a problem. I am just asking that if we have said that the limit exists, then we have also said that the derivative of g can not be zero around a? $\endgroup$ – user119615 Jan 29 '15 at 15:57
  • $\begingroup$ If the derivative of g was zero arbitrary close to a, then it would be a problem?, because then the limit value would not be defined? $\endgroup$ – user119615 Jan 29 '15 at 15:57
  • $\begingroup$ g' is not zero in an neighbourhood of a (except possibly for a itself) by assumption $\endgroup$ – Loreno Heer Jan 29 '15 at 15:59
  • $\begingroup$ Yes sanjab, I see that they have required that on the wikipedia article. But my question is that if you already say that $\lim_{x \rightarrow a}\frac{f'(x)}{g'(x)}$ exists, then you have allready said this(?), so you don't need to state it aswell as a separate point(?), as they have on Wikipedia? But problem 2 is my main concern. $\endgroup$ – user119615 Jan 29 '15 at 16:08

If $g$ had a zero in every neighborhood of $a$, then it would have a sequence of zeros tending to $a$, and by Rolle's theorem there would be a zero of $g'$ between each adjacent pair of them. This would contradict the assumption that there is an interval around $a$ where $g'$ has no zeros.

Some calculus textbook require (for simplicity, I guess) that a function must be defined in a punctured neighborhood of $a$ in order for it to have a limit, and if you use this definition it is automatic that $g'$ is nonzero near $a$ if $f'/g'$ has a limit. However, this restriction is usually not imposed in more advanced treatments of limits (where it is only required that any punctured neighborhood of $a$ contains some point where the function is defined), and then one must add the assumption that $g' \neq 0$ explicitly.

  • $\begingroup$ Thank you very much for your answer! $\endgroup$ – user119615 Jan 29 '15 at 16:22
  • $\begingroup$ You're welcome! $\endgroup$ – Hans Lundmark Jan 29 '15 at 16:26

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